Math 2280 - Exam 1
University of Utah
Spring 2014
Name: Solutions by Dylan Zwick
This is a 50 minute exam. Please show all your work, as a worked
problem is required for full points, and partial credit may be rewarded for
some work in the right direction.
1
1. (15 Points) Differential Equation Basics
(a) (5 points) What is the order of the differential equation given below?1
2
′′
x2 y 3y (4) − ex y 2 y ′ + 3xy = 42
Solution - Fourth Order.
(b) (5 points) Is the differential equation given below linear?
y (3) + y 3 = x + 3
Solution - No.
(c) (5 points) On what intervals are we guaranteed a unique solution
exists for the differential equation below?
(x + 1)y ′ + e2x y =
x2 + 2
x−1
Solution - We can rewrite the differential equation as:
y′ +
e2x
x2 + 2
y=
.
(x + 1)
(x + 1)(x − 1)
Where guaranteed a unique solution exists on any interval where
e2x
x2 + 2
both
and
are continuous. These are the
(x + 1)
(x + 1)(x − 1)
intervals (−∞, −1), (−1, 1), and (1, ∞).
1
Extra credit - Solve this differential equation! Just kidding. Do not attempt to solve
it.
2
2. (10 points) Phase Diagrams
Find the critical points for the autonomous equation:
ciP
=
kP(P
—
M)(H
—
P),
where k, M, ii > 0 and MT > I-I. Draw the corresponding phase
diagram, and indicate if the critical points are stable, unstable, or
semistable.
The critical points are the values of P for which
= 0.
These are the points P = 0, H, MT. The corresponding phase diagram
is:
Solution
-
-
U
/‘i
1-i
From the phase diagram we see that P
points, while P H is unstable.
3
=
0, II are stable critical
3. (20 Points) Separable Equations
Solve the initial value problem
dP
= 5P 2 ,
dt
where P (0) = 2.
Why would this be an example of a “doomsday” equation? According to this differential equation, when is “doomsday”?
Solution - We can rewrite this differential equation as
dP
= 5dt.
P2
Integrating both sides we get
−
1
= 5t + C.
P
Solving for P we get
P (t) =
1
.
C − 5t
1
Plugging in the initial condition P (0) = 2 we get C = , and so
2
P (t) =
2
.
1 − 10t
1
we have a
This is called a “doomsday” equation because at t =
10
1
vertical asymptote. So, t =
is “doomsday” for this equation.
10
4
4. (35 points) Linear Differential Equations
(a) (20 points) Solve the initial value problem
2
y ′ + 4xy − 3y + 3e−2x = 0,
y(0) = 4.
Solution - We can rewrite this equation as
2
y ′ + (4x − 3)y = −3e−2x .
This is a first-order linear differential equation. The integrating
factor is
R
ρ(x) = e
(4x−3)dx
= e2x
2 −3x
.
Multiplying both sides by this integrating factor we get
e2x
2 −3x
y ′ + (4x − 3)e2x
2 −3x
y=
d 2x2 −3x e
y = −3e−3x .
dx
Integrating both sides we get
e2x
2 −3x
y = e−3x + C
2
⇒ y(x) = e−2x + Ce−2x
2 +3x
.
Plugging in the initial condition we get
4 = y(0) = 1 + C ⇒ C = 3.
So, the unique solution to the initial value problem is
2
y(x) = e−2x + 3e−2x
5
2 +3x
.
(b) (15 points) Find the general solution to the differential equation
y ′′ − y ′ − 6y = 0.
Solution - The corresponding characteristic equation is
r 2 − r − 6 = (r − 3)(r + 2).
So, the roots are r = −2, 3, and the general solution to the differential
equation is
y(x) = c1 e3x + c2 e−2x .
6
5. (20 points) Euler’s Method
Use Euler’s method with step size h = 1 to estimate y(2), where y(x)
is the solution to the initial value problem
dy
= 2x + 3y,
dx
y(0) = 2.
Solution - Using Euler’s method our estimate for y(1) is
y(1) ≈ 2 + (2(0) + 3(2)) = 8.
Our estimate for y(2) is
y(2) ≈ 8 + (2(1) + 3(8)) = 34.
7